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NIELIT 2016 MAR Scientist C - Section A: 15
Lakshman Patel RJIT
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Quantitative Aptitude
Apr 2, 2020
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Nov 1, 2020
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soujanyareddy13
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The value of $\large\frac{(0.96)^3-(0.1)^3}{(0.96)^2+0.096+(0.1)^2}$ is :
$0.86$
$0.95$
$0.97$
$1.06$
nielit2016mar-scientistc
algebra
Lakshman Patel RJIT
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Quantitative Aptitude
Apr 2, 2020
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These type of questions are based on direct formulae of a^3 -b^3 and/or a^2 – b^2 etc.
Here, $a^{3}-b^{3} =(a-b)(a ^{2}+ b ^{2} +ab)$
So Simply, from question, a= 0.96 and b=0.10
So, Answer is 0.86
Devwritt
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Nov 19, 2020
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